(1000) Problem 9: The figure shows a cross section ofthree long, straight, and p

Question

(1000) Problem 9: The figure shows a cross section ofthree long, straight, and parallel current-carrying wires. Notice we are using the common convention that crosses denote vectors pointing into the page and dots denote vectors coming out of the page 5.00 A 10.0 A -10.0 cmI 20.0 A Otheexpertta.com > ? 50% Part (a) Find the magnitude of the force per unit length, in newtons per meter, on wire A. Grade Summary Deductions Potential 0% 100% sin0 cotan asin acos0 atan acotan) sinhO cosh tanh cotanh0 Submissions Attempts (000 per attempt) cos remaining detailed view END Degrees Radians NO Submit I give up! Hints: Feedback: ? ? 50% Part (b) Find the direction of the force for wire A. Measure this angle counterclockwise, in degrees, from the positive x-axis

Explanation / Answer

magnetic force per unit length between two wires carrying current I1 and I2 at a distance of d is given by:

mu*I1*I2/(2*pi*d)

force between wires carrying current in same direction is attractive in nature and vice versa.

coordinates of the points are:

B: (0,0)

C: (0.1,0)

A:(0.05,0.0866)

force on wire A due to wire B:

direction of current in A and current in B is in different direction.

so force is repulsive in nature.

force direction is from B to A.

vector along the force=(0.05,0.0866)-(0,0)=(0.05,0.0866)

unit unit vector along the force=(0.05,0.0866)/0.1=(0.5,0.866)

force magnitude per unit length=4*pi*10^(-7)*10*5/(2*pi*0.1)

=10^(-4) N/m

force in vector =F1=10^(-4)*(0.5,0.866) N/m

force on wire A due to wire C:

direction of current in A and current in C is in different direction.

so force is repulsive in nature.

force direction is from C to A.

vector along the force=(0.05,0.0866)-(0.1,0)=(-0.05,0.0866)

unit unit vector along the force=(-0.05,0.0866)/0.1=(-0.5,0.866)

force magnitude per unit length=4*pi*10^(-7)*20*5/(2*pi*0.1)

=2*10^(-4) N/m

force in vector =F2=2*10^(-4)*(-0.5,0.866) N/m

total force per unit length=F1+F2

=(-0.5,2.598)*10^(-4) N/m

magnitude of force per unit length

=sqrt(0.5^2+2.598^2)*10^(-4) N/m

=2.6457*10^(-4) N/m

part B:

unit vector along direction of force=(-0.5,2.598)/2.6457

=(-0.189,0.98197)

angle counter-clockwise from +ve x axis=180-arctan(0.98197/0.189)

=100.89 degrees

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